The number of solutions of the equation$\cos 6x+{\tan }^{2}x+\cos 6x\cdot {\tan }^{2}x=1$ , in the interval $[0,2\pi ]$ is _____

he number of solutions of the equation$\cos 6x+{\tan }^{2}x+\cos 6x\cdot {\tan }^{2}x=1$ , in the interval $[0,2\pi ]$ is _____

$\begin{array}{l}\cos 6x(1+{\tan }^{2}x)=1-{\tan }^{2}x\\ \Rightarrow \cos 6x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\\ \Rightarrow \cos 6x=\cos 2x\\ \Rightarrow 6x=2n\pi \pm 2x,n\in z\\ \Rightarrow 6x=2n\pi +2x,n\in z\left(or\right)6x=2n\pi -2x,n\in z\\ \Rightarrow 4x=2n\pi ,n\in z\left(or\right)8x=2n\pi ,n\in z\\ \Rightarrow x=\frac{n\pi }{2},n\in z\left(or\right)x=\frac{n\pi }{4},n\in z\\ \Rightarrow x=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi \left(or\right)x=0,\frac{\pi }{4},\frac{\pi }{2},\frac{3\pi }{4},\pi ,\frac{5\pi }{4},\frac{6\pi }{4},\frac{7\pi }{4},2\pi \\ \therefore x=0,\frac{\pi }{4},\frac{3\pi }{4},\pi ,\frac{5\pi }{4},\frac{7\pi }{4},2\pi (\because \tan x=\frac{\sin x}{\cos x},x\notin \frac{\pi }{2},\frac{3\pi }{2})\end{array}$

 $\therefore$ the number of solutions for the given equation are 7 $(\because x\in [0,2\pi ])$